Holds for b x <≤ξ. The limit as b x →- shows that this is also valid for .b x = Therefore φ is differentiable (to the left ) at b and )).(,()('b b f b φφ=
(b) It is sufficient to check that u satisfies the differential equation at b .The function u is differentiable to the left and the right at this point , and both derivatives are equal to ))(,(b b f φ.
We come now to the main theorem of this section.
VII. Existence and Uniqueness Theorem .Let )(D C f ∈ satisfy a local Lipschitz condition with respect to y in D ,where 2R D ? is open. Then for every D ∈),(ηξ the initial value problem ),,('y x f y = ηξ=)(y
Has a solution φ that cannot be extended and that to the left and to the right comes arbitrarily close to the boundary of D .The solution is uniquely determined in the sense that every solution of (7) is a restriction of φ .
Definition . The statement “φ comes arbitrarily close to the boundary of D to the right ” is defined as follows :If G is the closure of graph φ and if +G is the set of points G y x ∈),( with ξ≥x ,then
(a) +G is not a compact subset of D .
An equivalent formulation that gives a better understanding
reads as follows : φ exists to the right in an interval =∝<≤b b x (ξ is allowed ), and one of the following cases applies:
(b) =∝b ;the solution exists for all ξ≥x .
(c) <∝b and =∝-
→)(sup lim x b x φ ; the solution “becomes infinite.”
(d) <∝b and 0))(,(inf lim =-→
x x b x φρ ,where ),(y x ρ denotes the distance from the point ),(y x to the boundary of D ; the solution “comes arbitrarily close to the boundary of D ” Indeed , statement (a)says that +G is either unbounded
(case(b)or (c))or is bounded and contain boundary points of d(case(d)).
We have repeatedly encountered these three types of behavior .In the example x e y y sin '= of 1.VIII,(b)or (c) holds to the left and to the right ,depending on the value of )0(y .For the equation 1')2(-=y y in the upper half plane 0>y ,all solution are given by )(c x c x y ->+=.Here ,case (b) prevails to the right and case (d) to the left.
Proof . Uniqueness . We prove the statement ”if φ and ψ are two solutions of the initial value problem and if J is common interval of existence of both solutions with J ∈ξ,then
ψφ= in J ”.
Let us assume on the contrary that there exist ,say to the right of ξ ,points J x ∈ with )()(x x ψφ≠. Then there also exists a first point J x ∈0 to the right of ξ where the two solutions separate .This 0x is the largest number with the
property that )()(x x ψφ= for 0x x ≤≤ξ(ξ=0x is not excluded).
However ,we know from IV that there exists a local solution through the point ))(,(00x x φ and that it is uniquely determined .In other words, )()(x x ψφ= in a right neighborhood of 0x .This is a contradiction to our assumption about 0x .The uniqueness to the left is proved similarly.
Existence .By Theorem IV there exists a local solution to
(7),and as we have just proved ,the uniqueness statement (U) of V holds .Thus Corollary V guarantees the existence of a nonextendable solution φ,and we have only to show that it comes arbitrarily close to the boundary of D (we consider only the case “to the right ”in the direction of increasing ξ≥x x ,).
Assume that (a) is false .Then +G is a compact subset of
D , and φ exists in a finite interval b x <≤ξ or b x ≤≤ξ.In the first case ,Lemma VI.(a)can be applied ,i.e., φ can be extended to ][b ,ξ.In the second case ,D b b ∈))(,(φ,and there exists a local solution ψ that passes through this
point .Applying VI.(b),one again obtains an extension of φ. In either case ,we have a contradiction to the assumption that φ cannot be extended .This completes the proof of the theorem.
VIII . Exercise .Let ),,(z t x k be continuous for a x t ≤≤≤0 ,<∝∝<-z and satisfy a Lipschitz condition in z ,
z z L z t x k z t x k -≤-),,(),,(,
And let )(x g be continuous for a x ≤≤0.Show, by apply the fixed point theorem 5.IX,that the Volterra integral equation ”
dt t u t x k x g x u x
?+=0))(,,()()( Has exactly one continuous solution in a x ≤≤0.
IX . exercise .Prove :If ),(y x f satisfies a local Lipschitz condition with respect to y in the set 2R D ? and if D A ? is compact and f bounded on A ,then f satisfies a Lipschitz condition with respect to y in A .In particular ,if ][),(,b a C w v ∈and graph v ,graph D w ?,then there exists 0>L such that )()())(,())(,(x w x v L x w x f x v x f -≤- in ][b a ,.
X .Exercise .Prove :If f is continuous in the open set D and φ is a solution of (7) in the interval [),b ξ with <∝b that comes arbitrarily close to the boundary of D to the right,then
at least one of the following two cases applies(both can happen at the same time):
('c )∝+→)(x φ or ∝-as -→b x ;
('d )0))(,(→x x φρ as -→b x .
This sharpens the statement in VII.
Hint :Show :If b G is the intersection of graph with the line b x = ,then D G b ??(the boundary of D ).
XI .Exercise .Rosenblatt ’s Condition .Let the function ),(y x f be continuous in the strip ][a J R J S ,0,=?= and satisfy the condition z y x
q z x f y x f -≤-),(),( for a x ≤≤0 and R z y ∈, with 1 ),('y x f y = in J ,η=)0(y Has exactly one solution and that this solution can be obtained by method of successive approximations .The above condition was introduced by Rosenblatt(1909) . Hint .In the Banach space B of all function )(J C u ∈ with finite norm ? ?????≤≤=a x x x u u 0:)(sup :, The operator T , dt t u t f x Tu x ?+=0))(,(:))((η, Satisfies the Lipschitz condition (5.3).If u is a fixed point of T , then ημ+=y is a solution of the initial value problem. Supplement: Singular Initial Value Problems Here we consider a singular initial value problem for a differential equation of second order, ),('''y x f y x y =+α in (]b J ,00=,0)0(,)0('==y y η.(8) This problem is closely connected to the problem of finding rotationally symmetric solutions of the nonlinear elliptic equation ),(u r f u =?, Where n R x ∈ and x r =. 习题2.5 2.ydy x xdy ydx 2=- 。 解: 2x ,得: ydy x xdy ydx =-2 c y x y d +-=221 即c y x y =+2 2 1 4. xy x y dx dy -= 解:两边同除以x ,得 x y x y dx dy - =1 令u x y = 则dx du x u dx dy += 即 dx du x u dx dy +=u u -=1 得到 ()2ln 2 1 1y c u -=, 即2 ln 21?? ? ??-=y c y x 另外0=y 也是方程的解。 6.()01=-+xdy ydx xy 解:0=+-xydx xdy ydx x d x y x d y y d x -=-2 得到c x y x d +-=??? ? ??2 21 即 c x y x =+2 2 1 另外0=y 也是方程的解。 8. 32 x y x y dx dy += 解:令 u x y = 则: 21u x u dx du x u dx dy +=+= 即2 1u x dx du x = 得到22x dx u du = 故c x u +-=-11 即 21 1x x c y += 另外0=y 也是方程的解。 10. 2 1?? ? ??+=dx dy dx dy x 解:令 p dx dy = 即p p x 2 1+= 而 p dx dy =故两边积分得到 c p p y +-=ln 2 12 因此原方程的解为p p x 21+=,c p p y +-=ln 212 。 12.x y xe dx dy e =?? ? ??+-1 解: y x xe dx dy +=+1 习题2.2 求下列方程的解。 1.dx dy =x y sin + 解: y=e ?dx (?x sin e ?-dx c dx +) =e x [- 2 1e x -(x x cos sin +)+c] =c e x -21 (x x cos sin +)是原方程的解。 2.dt dx +3x=e t 2 解:原方程可化为: dt dx =-3x+e t 2 所以:x=e ?-dt 3 (?e t 2 e -? -dt 3c dt +) =e t 3- (5 1e t 5+c) =c e t 3-+5 1e t 2 是原方程的解。 3.dt ds =-s t cos +21t 2sin 解:s=e ?-tdt cos (t 2sin 2 1?e dt dt ?3c + ) =e t sin -(?+c dt te t t sin cos sin ) = e t sin -(c e te t t +-sin sin sin ) =1sin sin -+-t ce t 是原方程的解。 4. dx dy n x x e y n x =- , n 为常数. 解:原方程可化为:dx dy n x x e y n x += )(c dx e x e e y dx x n n x dx x n +??=?- )(c e x x n += 是原方程的解. 5. dx dy +1212--y x x =0 解:原方程可化为:dx dy =-1212+-y x x ?=-dx x x e y 1 2(c dx e dx x x +?-221) )21(ln 2+=x e )(1 ln 2?+--c dx e x x =)1(1 2 x ce x + 是原方程的解. 6. dx dy 234xy x x += 解:dx dy 234xy x x += =23y x +x y 令 x y u = 则 ux y = dx dy =u dx du x + 因此:dx du x u +=2u x 21u dx du = dx du u =2 c x u +=33 1 c x x u +=-33 (*) 将x y u =带入 (*)中 得:3433cx x y =-是原方程的解. 习题2.2 求下列方程的解 1. dx dy =x y sin + 解: y=e ?dx (?x sin e ?-dx c dx +) =e x [- 21 e x -(x x cos sin +)+c] =c e x -2 1 (x x cos sin +)是原方程的解。 2. dt dx +3x=e t 2 解:原方程可化为: dt dx =-3x+e t 2 所以:x=e ? -dt 3 (?e t 2 e -?-dt 3c dt +) =e t 3- (5 1 e t 5+c) =c e t 3-+5 1 e t 2 是原方程的解。 3. dt ds =-s t cos + 21t 2sin 解:s=e ? -tdt cos (t 2sin 2 1 ?e dt dt ? 3c + ) =e t sin -(?+c dt te t t sin cos sin ) = e t sin -(c e te t t +-sin sin sin ) =1sin sin -+-t ce t 是原方程的解。 4. dx dy n x x e y n x =- , n 为常数. 解:原方程可化为: dx dy n x x e y n x += )(c dx e x e e y dx x n n x dx x n +??=?- )(c e x x n += 是原方程的解. 5. dx dy + 1212 --y x x =0 解:原方程可化为: dx dy =-1212 +-y x x ? =-dx x x e y 2 1 2(c dx e dx x x +? -2 21) ) 2 1(ln 2 + =x e )(1ln 2 ?+- -c dx e x x =)1(1 2 x ce x + 是原方程的解. 6. dx dy 2 3 4xy x x += 解: dx dy 2 3 4 xy x x += =2 3y x + x y 令 x y u = 则 ux y = dx dy =u dx du x + 因此:dx du x u += 2 u x 2 1u dx du = dx du u =2 c x u +=3 31 c x x u +=-33 (*) 将 x y u =带入 (*)中 得:3 4 3 3cx x y =-是原方程的解. 常微分方程 2.1 1. xy dx dy 2=,并求满足初始条件:x=0,y=1的特解. 解:对原式进行变量分离得 。 故它的特解为代入得 把即两边同时积分得:e e x x y c y x x c y c y xdx dy y 2 2 ,11,0,ln ,21 2 =====+== ,0)1(.22 =++dy x dx y 并求满足初始条件:x=0,y=1的特解. 解:对原式进行变量分离得: 。 故特解是 时,代入式子得。当时显然也是原方程的解当即时,两边同时积分得;当x y c y x y x c y c y x y dy dx x y ++=====++=+=+≠=+- 1ln 11 ,11,001ln 1 ,11ln 0,1112 3 y xy dx dy x y 32 1++ = 解:原式可化为: x x y x x y x y x y y x y c c c c x dx x dy y y x y dx dy 2 2 2 2 22 2 2 3 22 3 2 )1(1)1)(1(),0(ln 1ln 21ln 1ln 2 1 1 1,0111=++ =++ ≠++-=+ +=+≠+ ? + =+) 故原方程的解为(即两边积分得故分离变量得显然 .0;0;ln ,ln ,ln ln 0 110000 )1()1(4===-==-+=-++=-=+≠===-++x y c y x xy c y x xy c y y x x dy y y dx x x xy x y xdy y ydx x 故原方程的解为即两边积分时,变量分离是方程的解,当或解:由: 10ln 1ln ln 1ln 1,0 ln 0 )ln (ln :931:8. cos ln sin ln 0 7ln sgn arcsin ln sgn arcsin 1 sgn 11,)1(,,,6ln )1ln(2 11 11,11,,,0 )()(:5332 2 22 2 22 2 22 2 c dx dy dx dy x y cy u d u u dx x x y u dx x y dy x y ydx dy y x x c dy y y y y dx dy c x y tgxdx ctgydy ctgxdy tgydx c x x x y c x x u dx x x du x dx du dx du x u dx dy ux y u x y y dx dy x c x arctgu dx x du u u u dx du x u dx du x u dx dy ux y u x y x y x y dx dy dx x y dy x y e e e e e e e e x y u u x y x u u x y x y y x x x +===+=+-===-?-=--+-=-=+-===-=+?=+?=?=--=+===-+=+-=++ =++-++=++===+-==-++-+-- 两边积分解:变量分离:。 代回原变量得:则有:令解:方程可变为:解:变量分离,得 两边积分得:解:变量分离,得::也是方程的解。 另外,代回原来变量,得两边积分得:分离变量得:则原方程化为: 解:令:。两边积分得:变量分离,得:则令解: 常微分方程(第三版) 答案 常微分方程习题答案 2.1 1.?Skip Record If...?,并求满足初始条件:x=0,y=1的特解. 解:对原式进行变量分离得 ?Skip Record If...??Skip Record If...?并求满足初始条件:x=0,y=1的特解. 解:对原式进行变量分离得: ?Skip Record If...?3 ?Skip Record If...? 解:原式可化为: ?Skip Record If...??Skip Record If...??Skip Record If...? ?Skip Record If...? 12.?Skip Record If...? 解?Skip Record If...??Skip Record If...? ?Skip Record If...? 15.?Skip Record If...? ?Skip Record If...?16.?Skip Record If...? 解:?Skip Record If...? ?Skip Record If...?,这是齐次方程,令?Skip Record If...? 17. ?Skip Record If...? 解:原方程化为?Skip Record If...? 令?Skip Record If...? 方程组?Skip Record If...??Skip Record If...? 则有?Skip Record If...? 令?Skip Record If...? 当?Skip Record If...?当?Skip Record If...? 另外 ?Skip Record If...? ?Skip Record If...? 习题3.1 1 求方程dx dy =x+y 2通过点(0,0)的第三次近似解; 解: 取0)(0=x ? 20020012 1)()(x xdx dx y x y x x x ==++=??? 522200210220 121])21([])([)(x x dx x x dx x x y x x x +=+=++=???? dx x x x y x x ])20 121([)(252003+++=?? = 118524400 1160120121x x x x +++ 2 求方程dx dy =x-y 2通过点(1,0)的第三次近似解; 解: 令0)(0=x ? 则 20020012 1)()(x xdx dx y x y x x x ==-+=??? 522200210220 121])21([])([)(x x dx x x dx x x y x x x -=-=-+=???? dx x x x y x x ])20 121([)(252003--+=?? =118524400 1160120121x x x x -+- 3 题 求初值问题: ?????=--=0 )1(22y y x dx dy R :1+x ≤1,y ≤1 的解的存在区间,并求解第二次近似解,给出在解的存在空间的误差估计; 解: 因为 M=max{22y x -}=4 则h=min(a,M b )=4 1 则解的存在区间为0x x -=)1(--x =1+x ≤4 1 令 )(0X ψ=0 ; )(1x ψ=y 0+?-x x x 0)0(2dx=31x 3+31; )(2x ψ =y 0+])3131([2132?-+-x x x dx=31x 3-9x -184x -637x +4211 又 y y x f ??),(2≤=L 则:误差估计为:)()(2x x ψ-ψ≤32 2 )12(*h L M +=2411 4 题 讨论方程:31 23y dx dy =在怎样的区域中满足解的存在唯一性定理的条件, 并求通过点(0,0)的一切解; 解:因为y y x f ??),(=3221-y 在y 0≠上存在且连续; 而312 3y 在y 0 σ≥上连续 由 3123y dx dy =有:y =(x+c )23 又 因为y(0)=0 所以:y =x 2 3 另外 y=0也是方程的解; 故 方程的解为:y =?????≥00023 x x x 或 y=0; 6题 证明格朗瓦耳不等式: 设K 为非负整数,f(t)和g(t)为区间βα≤≤t 上的连续非负函数, 习题1.2 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2 另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 解:y 2dx=-(x+1)dy 2y dy dy=-1 1+x dx 两边积分: - y 1 =-ln|x+1|+ln|c| y=|)1(|ln 1+x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 321++ 解:原方程为:dx dy =y y 21+3 1 x x + y y 21+dy=3 1 x x +dx 两边积分:x(1+x 2 )(1+y 2 )=cx 2 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1 +dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- 令 x y =u 则dx dy =u+x dx du 代入有: -1 12++u u du=x 1dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2x y . 6. x dx dy -y+22y x -=0 解:原方程为: dx dy =x y +x x | |-2)(1x y - 则令 x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32 +=0 解:原方程为:dx dy =y e y 2 e x 3 2 e x 3-3e 2 y -=c. 9.x(lnx-lny)dy-ydx=0 解:原方程为: dx dy =x y ln x y 习题2.2 求下列方程的解 1.dx dy =x y sin + 解: y=e ?dx (?x sin e ?-dx c dx +) =e x [- 2 1e x -(x x cos sin +)+c] =c e x -21 (x x cos sin +)是原方程的解。 2.dt dx +3x=e t 2 解:原方程可化为: dt dx =-3x+e t 2 所以:x=e ?-dt 3 (?e t 2 e -? -dt 3c dt +) =e t 3- (5 1e t 5+c) =c e t 3-+5 1e t 2 是原方程的解。 3.dt ds =-s t cos +21t 2sin 解:s=e ?-tdt cos (t 2sin 2 1?e dt dt ?3c + ) =e t sin -(?+c dt te t t sin cos sin ) = e t sin -(c e te t t +-sin sin sin ) =1sin sin -+-t ce t 是原方程的解。 4. dx dy n x x e y n x =- , n 为常数. 解:原方程可化为:dx dy n x x e y n x += )(c dx e x e e y dx x n n x dx x n +??=?- )(c e x x n += 是原方程的解. 5. dx dy +1212--y x x =0 解:原方程可化为:dx dy =-1212+-y x x ?=-dx x x e y 21 2(c dx e dx x x +?-221) )21(ln 2+=x e )(1 ln 2?+--c dx e x x =)1(1 2 x ce x + 是原方程的解. 6. dx dy 234xy x x += 解:dx dy 234xy x x += =23y x +x y 令 x y u = 则 ux y = d x d y =u dx du x + 因此:dx du x u +=2u x 21u dx du = dx du u =2 c x u +=33 1 c x x u +=-33 (*) 将x y u =带入 (*)中 得:3433cx x y =-是原方程的解. 常微分方程第三版答 案 习题1.2 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 解:y 2dx=-(x+1)dy 2y dy dy=-1 1+x dx 两边积分: - y 1 =-ln|x+1|+ln|c| y=|)1(|ln 1+x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 321++ 解:原方程为:dx dy =y y 21+3 1 x x + y y 21+dy=3 1 x x +dx 两边积分:x(1+x 2)(1+y 2)=cx 2 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1 +dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- 令 x y =u 则dx dy =u+x dx du 代入有: -1 12++u u du=x 1dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2 x y . 6. x dx dy -y+22y x -=0 解:原方程为: dx dy =x y +x x | |-2)(1x y - 则令 x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32 +=0 解:原方程为:dx dy =y e y 2 e x 3 2 e x 3-3e 2 y -=c. 习题1.2 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2 另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2 dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 解:y 2dx=-(x+1)dy 2 y dy dy=-1 1+x dx 两边积分: - y 1 =-ln|x+1|+ln|c| y=|)1(|ln 1+x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 321++ 解:原方程为:dx dy =y y 21+31 x x + y y 21+dy=31 x x +dx 两边积分:x(1+x 2 )(1+y 2 )=cx 2 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1 +dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- 令 x y =u 则dx dy =u+x dx du 代入有: -1 12++u u du=x 1dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2 x y . 6. x dx dy -y+22y x -=0 解:原方程为: dx dy =x y +x x | |-2)(1x y - 则令 x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32+=0 解:原方程为:dx dy =y e y 2 e x 3 2 e x 3-3e 2 y -=c. 9.x(lnx-lny)dy-ydx=0 解:原方程为: dx dy =x y ln x y 令x y =u ,则dx dy =u+ x dx du \ 习题 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2 另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2 dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 & 解:y 2 dx=-(x+1)dy 2y dy dy=-1 1+x dx 两边积分: - y 1 =-ln|x+1|+ln|c| y=|)1(|ln 1+x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 321++ 解:原方程为:dx dy =y y 21+3 1 x x + y y 21+dy=3 1 x x +dx 两边积分:x(1+x 2)(1+y 2)=cx 2 、 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1 +dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- { 令 x y =u 则dx dy =u+x dx du 代入有: -1 12++u u du=x 1dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2x y . 6. x dx dy -y+2 2y x -=0 解:原方程为: dx dy =x y +x x | |-2)(1x y - 则令 x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx } arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32 +=0 解:原方程为:dx dy =y e y 2 e x 3 ` 2 e x 3-3e 2 y -=c. (lnx-lny)dy-ydx=0 常微分方程第三版课后习题答案 习题1.2 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 解:y 2dx=-(x+1)dy 2 y dy dy=-11+x dx 两边积分: -y 1=-ln|x+1|+ln|c| y= | )1(|ln 1 +x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 3 2 1++ 解:原方程为:dx dy =y y 21+3 1 x x + y y 2 1+dy=3 1x x +dx 两边积分:x(1+x 2)(1+y 2)=cx 2 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1+dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- 令x y =u 则 dx dy =u+x dx du 代入有: -1 12++u u du=x 1 dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2x y . 6. x dx dy -y+22y x -=0 解:原方程为: dx dy =x y +x x ||-2)(1x y - 则令x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32+=0 习题 1. dx dy =2xy,并满足初始条件:x=0,y=1的特解。 解: y dy =2xdx 两边积分有:ln|y|=x 2+c y=e 2 x +e c =cex 2 另外y=0也是原方程的解,c=0时,y=0 原方程的通解为y= cex 2,x=0 y=1时 c=1 特解为y= e 2 x . 2. y 2dx+(x+1)dy=0 并求满足初始条件:x=0,y=1的特解。 解:y 2dx=-(x+1)dy 2y dy dy=-1 1+x dx 两边积分: - y 1 =-ln|x+1|+ln|c| y=|)1(|ln 1+x c 另外y=0,x=-1也是原方程的解 x=0,y=1时 c=e 特解:y= | )1(|ln 1 +x c 3.dx dy =y x xy y 321++ 解:原方程为:dx dy =y y 21+3 1 x x + y y 21+dy=3 1 x x +dx 两边积分:x(1+x 2 )(1+y 2 )=cx 2 4. (1+x)ydx+(1-y)xdy=0 解:原方程为: y y -1dy=-x x 1 +dx 两边积分:ln|xy|+x-y=c 另外 x=0,y=0也是原方程的解。 5.(y+x )dy+(x-y)dx=0 解:原方程为: dx dy =-y x y x +- 令 x y =u 则dx dy =u+x dx du 代入有: -1 12++u u du=x 1dx ln(u 2+1)x 2=c-2arctgu 即 ln(y 2+x 2)=c-2arctg 2x y . 6. x dx dy -y+22y x -=0 解:原方程为: dx dy =x y +x x | |-2)(1x y - 则令 x y =u dx dy =u+ x dx du 2 11u - du=sgnx x 1 dx arcsin x y =sgnx ln|x|+c 7. tgydx-ctgxdy=0 解:原方程为: tgy dy =ctgx dx 两边积分:ln|siny|=-ln|cosx|-ln|c| siny= x c cos 1=x c cos 另外y=0也是原方程的解,而c=0时,y=0. 所以原方程的通解为sinycosx=c. 8 dx dy +y e x y 32 +=0 解:原方程为: dx dy =y e y 2 e x 3 2 e x 3-3e 2 y -=c. (lnx-lny)dy-ydx=0 解:原方程为: dx dy =x y ln x y2.5常微分方程课后答案(第三版)王高雄
常微分方程课后答案(第三版)王高雄
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